Mathematics · Parabola

JEE Main 2025 — 2 April, Evening Shift — Question 34

Let the point $P$ of the focal chord $P Q$ of the parabola $y^{2}=16 x$ be $(1,-4)$ . If the focus of the parabola divides the chord $P Q$ in the ratio $m: n, g c d(m, n)=$ 1 , then $m^{2}+n^{2}$ is equal to:

Option A:
17
Correct
Option B:
26
Option C:
10
Option D:
37

Answer: A

Step-by-step solution

$P\left(a t^{2}, 2 a t\right) \equiv P\left(4 t^{2}, 8 t\right) \equiv(1,-4)$

$8 t=-4 \Rightarrow t=-\frac{1}{2}$

$\therefore Q\left(\frac{a}{t^{2}}, \frac{-2 a}{t}\right) \quad\left(t_{1} \cdot t_{2}=-1\right)$

$S(4,0)$ is the focus

$P S=a+a t^{2}$

$Q S=a+\frac{a}{t^{2}}=\frac{a t^{2}+a}{t^{2}}$

$\frac{P S}{Q S}=t^{2}=\frac{1}{4}=\frac{m}{n}$

$\therefore \quad m^{2}+n^{2}=17$

Answer key and solution verified before publishing.

Practise Parabola

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2025
Paper: 2 April, Evening Shift
Subject: Mathematics
Chapter: Parabola
Topic: Chords connected with a Parabola